TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Equivalent characterisations of a normal subgroup by conjugates and left and right cosets

Statement

Let $N \leq G$ . The following conditions are equivalent:

$N ⊴ G$ (Normal subgroup: invariance under conjugation);
$g N g^{- 1} \subseteq N$ for every $g \in G$ ;
$g N = N g$ for every $g \in G$ , where these are the left and right cosets of $N$ represented by $g$ .

Facts & Assumptions

Given: A group $G$ and a subgroup $N \leq G$ .

[F1]

The subgroup $N$ is normal when $g N g^{- 1} = N$ for every $g \in G$ (Normal subgroup: invariance under conjugation).

[L1]

In a group, $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ (In a group $e^{- 1} = e$ , $(g^{- 1})^{- 1} = g$ and $(g h)^{- 1} = h^{- 1} g^{- 1}$ , the order of the last product being essential).

Proof

technique · direct

1.1F1

Condition 1 implies condition 2 because equality implies containment.

1.2givenL1algebra

Suppose condition 2 holds. Applying it to $g^{- 1}$ gives $g^{- 1} N g \subseteq N$ ; conjugating this containment by $g$ and using $(g^{- 1})^{- 1} = g$ gives $N \subseteq g N g^{- 1}$ , while condition 2 gives the reverse containment. Hence $g N g^{- 1} = N$ for every $g$ , so condition 1 holds.

1.3givenF1L1algebra

Suppose condition 1 holds. If $x \in g N$ , then $x = g n = (g n g^{- 1}) g$ for some $n \in N$ , and $g n g^{- 1} \in N$ by [F1], so $x \in N g$ . Replacing $g$ by $g^{- 1}$ gives the reverse inclusion, hence $g N = N g$ .

1.4givenalgebra

Suppose condition 3 holds. For $n \in N$ , the element $g n$ lies in $g N = N g$ , so $g n = n^{'} g$ for some $n^{'} \in N$ ; therefore $g n g^{- 1} = n^{'} \in N$ . Thus $g N g^{- 1} \subseteq N$ and condition 2 holds.

2.1step 1.1step 1.2step 1.3step 1.4∎

Steps 1.1 through 1.4 prove that conditions 1, 2, and 3 are equivalent.

Depends on

Dependency tree · 2 levels

Sources

Encyclopedia of Mathematics, Normal subgroup